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PRACTICAL EXAMPLES.

To find the hypo

1. In the right-angled spherical triangle ABC.
The angle A=47°.54'.20"
The angle c=61°.50′.29")

Given

Answer. Ac=61°.4'.56", acute. (S. 212.)

thenuse AC.

2. In the right-angled spherical triangle ABC.
The angle A=31°51′

Given {The angle c=104.8'

Required the hypothenuse AC.

Answer. AC 113.55', obtuse. (S. 212.)

CHAP. VII.

II. PRACTICAL RULES FOR SOLVING THE DIFFERENT cases of RECTILATERAL, OR QUADRANTAL, SPHERICAL TRIANGLES, WITH THEIR APPLICATION BY LOGARITHMS.

(L) 1. When one side of a spherical triangle is 90°, or a quadrant, it is called a quadrantal triangle.

2. If two sides of a quadrantal triangle be each 90°, the triangle will be isosceles, each of the angles at the base will be 90°, and the base itself will be the measure of the vertical angle. If the three sides be each 90°, the three angles will be each 90°.

3. A quadrantal spherical triangle may be changed into a right-angled spherical triangle, and the contrary. (X. 138.)

(M) RULE I.

Subtract the angle opposite to the quadrantal side from 180°, and call the remainder the hypothenuse of a new triangle. The angle opposite to the quadrantal side must then be considered as a right angle, and the two remaining angles must be represented by the sides of the quadrantal triangle which are opposite to them.

The sides and angles of the right-angled spherical triangle, will represent the angles and sides of the quadrantal triangle.

A

B

Thus in the annexed figure, where AC is supposed to be a quadrant, the triangle ABC may be considered as a right-angled triangle, whose hypothenuse AC=180°-B, the base AB= Zc; and the perpendicular BC=the A; et contra.

(N) RULE II.

1. In the triangle ABC, if one of the sides ac be a quadrant, and another side BC less than a quadrant.

Produce the side Bc till it becomes a quadrant, and Adb will be a right-angled triangle; wherein AD is the measure of the angle C, DB is the complement of BC, and AE is the hypothenuse.

2. In the triangle AEC, where one of the sides AC is a quadrant, and another side ce greater than a quadrant.

Subtract 90° from CE, and the remainder will be the side DE of the right-angled triangle ADE; AE is the hypothenuse, and AD is the measure of the angle c.

The same reasoning will apply whether AB and AE be less or greater than quadrants, for in either case ADB and ADE will be right-angled triangles.

(O) CASE I. Given a quadrantal side, its adjacent and opposite angle, to find the rest.

The quadrantal side ac=90°.00'

Given Its adjacent /C=

Its opposite B=

42°.12'

115°.20'

Required the sides AB, BC and the ▲ A.

BY RULE II. (N. 228.)

Because AC and CD are quadrants, the angles CAD and CDA are each of them a right angle (L. 227.), and AD is the measure of the angle c.

The angles ABC and ABD, being supplements to each other, may be considered as the same angle, since they have the same sine, tangent,

&c.

1. To find AB, by Case XII. right-angled spherics. sine ABD : sine AD=sine c::rad: sine ab.

Here AB is acute=48°.

2. To find BD and hence BC, by Case X. right-angled spherics. rad cot ABD::tang AD=tang/c: sine BD. Here BD is acute=25°.25', and its complement=BC=64°.35′. 3. To find the BAD, by Case XI. right-angled spherics. COS AD COSC: rad:: cos ABD=cos 64°.40′ sine BAD. Here BAD is acute=35°.17', and its complement BAC=54°.43′. BY RULE I. (M. 227.)

Here the supplement of the angle B= 64°.40', must represent the hypothenuse ac, the quadrantal side AC the right angle b. The angles c and a must represent the sides ab and bc, and the sides BC and AB must represent the angles a and c.

DA

B

a

1. To find the Lc, in the triangle abc, by Case V. rightangled spherics.

sine ac sine suppt. Z B: rad::sin ab=sinc: sin c=sin AB. Here thec or AB is acute=48°.

2. To find the La, in the triangle abc, by Case IV. rightangled spherics.

rad: cot accot suppt. B:: tangab=tang /c: cos a=cos BC. Here cosine of La=64°.35' acute.

3. To find bc, in the triangle abc, by Case VI. right-angled spherics.

cos ab=cos ≤ c:rad::cosac=cos suppt. B: cosbc=cos ▲ a. Here bc is acute=54°.43'.

BY CONSTRUCTION. (Plate V. Fig. 13.)

1. With the chord of 60 degrees describe the primitive circle; and through the centre P draw arD, and cre at right angles to it.

2. Set one foot of your compasses on 90 degrees on the line of semi-tangents, extend the other towards the beginning of the scale, till the degrees between them be equal to the angle c=42°.12′, and apply this extent from D to A.

3. Through the three points cae draw a great circle, then AC will represent the quadrantal side.

4. With the tangent of the complement of the angle B, and centre c, describe an arc; with the secant of the complement of the same angle, and centre A, cross it in o: with o as a centre, and radius oA, describe the great circle Bab. Then ABC is the triangle required.

To measure the required parts.

The sides AB and BC (C. 163.) will be 48° and 64°.35',
And the angle A=54°.43'. (G. 164.)

NOTE. The right-angled spherical triangle abc (Plate V. fig. 14.) into which the quadrantal triangle ABC is transformed, be constructed and measured exactly in the same manner as Case IV. of right-angled spherical triangles was constructed and measured. See Plate V. fig. 9.

may

(P) CASE II. Given a quadrantal side, and the other two sides, to find the rest.

The quadrantal side Ac= 90°†

Given

The side AB

The side BC

=115°. 9'

=113°.18'

Required the angles A, B, and c.

This example was formed from Example 2, Case VI. of right-angled spherics. The B was made equal to the supplement of the hypothenuse, the two sides ab and BC were made equal to the angles of the right-angled triangle; and hence the legs of the right-angled triangle become angles in the quadrantal triangle. In a similar manner the other examples were made.

BY RULE II. (N. 228.)

Make CD AC, then CAD, CDA, and adb are each of them a right angle, and AD measures the angle c (L. 227, and M. 124.); hence, if we take 90° from BC we get 23°.18' BD; and ADB is a right-angled triangle.

1. To find the LB, by Case IV. right-angled spherics.
rad cot AB::tang BD: COSB=78°.20′.
But B is obtuse (O. 211), and therefore 101°.40'.
2. To find AD, by Case VI. right-angled spherics.
Cos BD rad:: cos AB COS AD=62°.26'.

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But AD is obtuse (O. 211), and therefore=117°.34′.
3. To find the L DAB, by Case V. right-angled spherics.
Sine AB rad::sine BD sine / DAB=25°.55'.
Here DAB is acute (0.211.); and ▲ DAB+ 4 CAD
Z CAB 25°.55′+90°115°.55'.

BY RULE I. (M. 227.)

Here the quadrantal side ac must represent the right-angle b; AB the LC; and BC the a; ac the supplement of the B, ab the 4 c, and be the A.

A

1. To find the hypoth. ac, by Case XVI. right-angled spherics. Rad: cot/a-cot BC:: cot/c=cot AB: cosąc=sup ▲ B=78°.20′. Here ac or suppt. 2 B is acute (S. 212.); therefore B is obtuse

=101°.40'.

2. To find ab, by Case XV. right-angled spherics. Sine A sin BC: rad::cos c=cos AB: cos ab=cos ≤ c=62°.26'. Here abc is obtuse (S. 212.), and therefore=117°.34′.

3. To find bc, by Case XV. right-angled spherics. Sine csin AB: rad:: cosa cos BC: cos bc=cos / a=64°.5′. Here be A is obtuse (S. 212.), and therefore=115°.55'.

If the two preceding cases be thoroughly understood, there can be no difficulty in solving those which follow; for which reason, they are given as practical exercises.

(Q) CASE III. Given a quadrantal side, and its two adjacent angles, to find the rest.

The quadrantal side Ac=90°

Given

The angle

The angle

Answer. AB=48°, BC

540.43 Required AB, BC, and the angle B.

A=540.43′

c=42°.12

64°.35', and the angle B-115°.20'.

(R) CASE IV. Given a quadrantal side, one of the other sides, and the angle comprehended between them, to find the

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gles B and C. c=117°.34'.

Answer. BC=1130.18', the angle B-101°.40',

(S) CASE V. Given a quadrantal side, its adjacent angle, and a side opposite to that angle, to find the rest.

The quadrantal side Ac=90°

Given Its adjacent angle C=42°,12′

And the opposite side AB 48°.00'

Required BC, and the an

gles A and B.

Answer. BC=64°.35′, the angle A=54°.43′. and B=115°.20'. But the required parts are ambiguous, and therefore are either acute or obtuse.

(T) CASE VI. Given the quadrantal side, one of the other sides and an angle opposite to the quadrantal side, to find the rest.

The quadrantal side AC= 90°

Given The side AB

The angle B

=115°. 9'

Required BC, and the an

=101°.40'

gles A and c.

Answer. BC113°.18', the angle ▲ 115°.55', and c≈ 1170.34'.

CHAP. VIII.

OF

III. PRACTICAL RULES FOR SOLVING ALL THE CASES OBLIQUE-ANGLED SPHERICAL TRIANGLES, WITH A PERPENDICULAR; AND THEIR APPLICATION BY LOGARITHMS.

RULE I.

(U) The rule Prop. xxv. (X. 173.) will solve ten cases; see also the rules P. 178.

RULE II.

(W) When the three sides are given, to find the angles. Any one of the three sides may be called the base. Then, The tangent of half the base,

Is to the tangent of half the sum of the sides,

As the tangent of half the difference between the sides, Is to the tangent of the distance of a perpendicular from the middle of the base. (S. 179.)

According as this distance is less or greater than half the

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